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The graph relates. In `K_(eq.) vs.(1)/(T)` for a reaction. The reaction must be: A. exothermicB. endothermicC. `DeltaH` is negligibleD. high spontaneous at ordinary temperature |
Answer» In `(K_(2))/(K_(1))=(DeltaH)/(R )[(T_(2)-T_(1))/(T_(1)T_(2))]` Since `K` increase with decrease of temperature and thus `DeltaH=-ve`. |
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